Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 mm$. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

What are the diameter and cross-sectional area of the wire measured using the screw gauge?

There are two Vernier calipers both of which have $1 \mathrm{~cm}$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $\left(C_1\right)$ has $10$ squal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $\left(C_2\right)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $\mathrm{cm}$ ) by calipers $C_1$ and $C_2$, respectively, are

Student $A$ and Student $B$ used two screw gauges of equal pitch and $100$ equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is $0.322\, {cm}$. The absolute value of the difference between the final circular scale readings observed by the students $A$ and $B$ is …. .

[Figure shows position of reference $'O'$ when jaws of screw gauge are closed]

Given pitch $=0.1 \,{cm}$.

The circular scale of a micrometer has $200$ divisions and pitch of $2\,mm$ . Find the measured value of thickness of a thin sheet ………. $mm$

A screw gauge has $50$ divisions on its circular scale. The circular scale is $4$ units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of $0.5\, mm$ is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively